\newproblem{lay:2_9_27}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.9.27}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Suppose vectors $\mathbf{b}_1$, $\mathbf{b}_2$, ..., $\mathbf{b}_p$ span a subspace $W$, and let $\{\mathbf{a}_1, \mathbf{a}_2, ..., \mathbf{a}_q\}$ by any set
	in $W$ containing more than $p$ vectors. Fill in the details of the following argument to show that $\{\mathbf{a}_1, \mathbf{a}_2, ..., \mathbf{a}_q\}$
	must be linearly dependent. First, let $B=\begin{pmatrix}\mathbf{b}_1 & \mathbf{b}_2 & ...& \mathbf{b}_p\end{pmatrix}$ and
	$A=\begin{pmatrix} \mathbf{a}_1 &\mathbf{a}_2 & ...& \mathbf{a}_q\end{pmatrix}$.
	\begin{enumerate}[a.]
		\item Explain why for each vector $\mathbf{a}_j$, there exists a vector $\mathbf{c}_j$ in $\mathbb{R}^p$ such that $\mathbf{a}_j=B\mathbf{c}_j$.
		\item Let $C=\begin{pmatrix} \mathbf{c}_1 &\mathbf{c}_2 & ...& \mathbf{c}_q\end{pmatrix}$. Explain why there is a non-zero vector $\mathbf{u}$ such that
		      $C\mathbf{u}=\mathbf{0}$.
		\item Use $B$ and $C$ to show that $A\mathbf{u}=\mathbf{0}$. This shows that the columns of $A$ are linearly dependent.
	\end{enumerate}
}{
  % Solution
	\begin{enumerate}[a.]
		\item Each vector $\mathbf{a}_j$ is in $W$, that is spanned by the $\mathbf{b}_i$ vectors. That means that there exist some coefficients $c_{ji}$ such that \\
		      \begin{center}
						$\mathbf{a}_j=c_{j1}\mathbf{b}_1+c_{j2}\mathbf{b}_2+...+c_{jp}\mathbf{b}_p$
					\end{center}
					or what is the same \\
		      \begin{center}
						$\mathbf{a}_j=B\mathbf{c}_j$
					\end{center}
		\item Note that the $\mathbf{c}_j$ vectors are in $\mathbb{R}^p$ since they have $p$ components. The problem stated that $q>p$, that is there are more 
		      $\mathbf{c}_j$ vectors than $p$ (their dimension). By Theorem 6.2 of Chapter 2, we have that this set of equations is linearly dependent, that is, there exist
					some coefficients (not all of them zero) such that
		      \begin{center}
						$u_1\mathbf{c}_1+u_2\mathbf{c}_2+...+u_p\mathbf{c}_p=\mathbf{0}$
					\end{center}
					or
		      \begin{center}
						$C\mathbf{u}=\mathbf{0}$
					\end{center}
		\item Let us calculate $A\mathbf{u}$. From point a, we know that $A=BC$, therefore
		      \begin{center}
						$A\mathbf{u}=(BC)\mathbf{u}=B(C\mathbf{u})=B\mathbf{0}=\mathbf{0}$
					\end{center}
	\end{enumerate}
}
\useproblem{lay:2_9_27}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
